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Câu a : \(4x^2+4xy+y^2=\left(2x+y\right)^2\)
Câu b : \(9m^2+n^2-6mn=\left(3m-n\right)^2\)
Câu c : \(16a^2+25b^2+40ab=\left(4a+5b\right)^2\)
Câu d : \(x^2-x+\dfrac{1}{4}=\left(x-\dfrac{1}{2}\right)^2\)
\(a,4x^2+4xy+y^2=\left(2x\right)^2+4xy+y^2=\left(2x+y\right)^2\)
\(b,9m^2+n^2-6mn=\left(3m\right)^2-6mn+n^2=\left(3m-n\right)^2\)
\(c,16a^2+25b^2+40ab=\left(4a\right)^2+40ab+\left(5b\right)^2=\left(4a+5b\right)^2\)
@Yukru ơi! giúp câu D với!
Chúc bạn học tốt!
a) \(4x^2+4xy+y^2=\left(2x\right)^2+2.2x.y+y^2=\left(2x+y\right)^2\)
b) \(9m^2+n^2-6mn=\left(3m\right)^2-2.3m.n+n^2=\left(3m-n\right)^2\)
c) \(16a^2+25b^2+40ab=\left(4a\right)^2+2.4a.5b+\left(5b\right)^2=\left(4a+5b\right)^2\)
d) \(x^2-x+\dfrac{1}{4}=\left(x^2-2.\dfrac{1}{2}.x+\left(\dfrac{1}{2}\right)^2\right)=\left(x-\dfrac{1}{2}\right)^2\)
Bài 62: 25x2y6-60xy4z2+36y2z4=(5xy3)2-2.5xy3.(6yz2)2
Bài 63: 1/9u4v6-1/3u5v4+(1/2u3v)=(1/3u2v3)-2.1/3u2v3.1/2u2v3+(1/2u3v)
a) 4x2+4xy+y2
=(2x)2+2(2x)(y)+y2
=(2x+y)2
b)9m2+n2-6mn
=(3m)2-2(3m)n+n2
=(3m-n)2
c)16a2+25b2+40ab
=(4a)2+(5b)2+2(4a)(5b)
=(4a+5b)2
d)x2-x+\(\frac{1}{4}\)
=x2-2x.\(\frac{1}{2}\)+\(\left(\frac{1}{2}\right)^2\)
=\(\left(x-\frac{1}{2}\right)^2\)
a.
\(\frac{x^2}{4}+x+3=\frac{x^2}{4}+x+1+2=\left(\frac{x}{2}+1\right)^2+2>0;\forall x\)
b.
\(A=-3x^2+2x-5=-3\left(x^2-2.\frac{1}{3}x+\frac{1}{9}\right)-\frac{14}{3}=-3\left(x-\frac{1}{3}\right)^2-\frac{14}{3}\le-\frac{14}{3}\)
\(A_{max}=-\frac{14}{3}\) khi \(x=\frac{1}{3}\)
c.
Đề thiếu (để ý 2 số hạng cuối)
\(A=x^4-2x^3+x^2+3x^2-6x+3-1\)
\(=\left(x^2-x\right)^2+3\left(x-1\right)^2-1\ge-1\)
\(A_{min}=-1\) khi \(x=1\)
d.
\(27x^2-\frac{9}{2}x+\frac{3}{16}=3\left(9x^2-\frac{3}{2}x+\frac{1}{16}\right)=3\left(3x-\frac{1}{4}\right)^2\)
e.
\(=\left[\left(b+c\right)+a\right]^2+\left[\left(b+c\right)-a\right]^2+\left[a-\left(b-c\right)\right]^2+\left[a+\left(b-c\right)\right]^2\)
\(=2\left(b+c\right)^2+2a^2+2a^2+2\left(b-c\right)^2\)
\(=4a^2+2b^2+4bc+2c^2+2b^2-4bc+2c^2\)
\(=4\left(a^2+b^2+c^2\right)\)
f.
\(\left(a^2+b^2\right)\left(c^2+d^2\right)=a^2c^2+b^2d^2+a^2d^2+b^2c^2\)
\(=\left(a^2c^2+b^2d^2+2ac.bd\right)+\left(a^2d^2+b^2c^2-2ad.bc\right)\)
\(=\left(ac+bd\right)^2+\left(ad-bc\right)^2\)
a, \(4x^2+4xy+y^2=\left(4x\right)^2+2.2x.y+y^2\)
\(=\left(4x+y\right)^2\)
b, \(9m^2+n^2-6mn=\left(3m\right)^2-2.3m.n+n^2\)
\(=\left(3m-n\right)^2\)
c, \(16a^2+25b^2+40ab=\left(4a\right)^2+2.4a.5b+\left(5b\right)^2\)
\(=\left(4a+5b\right)^2\)
d, \(x^2-x+\dfrac{1}{4}=x^2-2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\)
\(=\left(x-\dfrac{1}{2}\right)^2\)
Chúc bạn học tốt!!!
bạn phải tách từng câu ra. chứ kiểu này k ai trả lời cho đâu
2)
a)x2-y2=(x+y).(x-y)=(87+13).(87-13)=100.74=7400
b)x3-3x2+3x-1=(x-1)3=(101-1)3=1003=1000000
c)x3+9x2+27x+27=(x+3)3=(97+3)3=1003=1000000
4)
a)x2-6x+10=x2-6x+9+1=(x-3)2+1>=1>0 voi moi x
b)4x-x2-5= -(x2-4x+5)= -(x2-4x+4+1)= -(x-2)2 - 1<0 voi moi x
\(a,x^2-x+\frac{1}{4}=\left(x-\frac{1}{2}\right)^2 \)
b,\(a^2-\frac{4}{3}ba+\frac{4}{9}b^2\)=\(\left(a-\frac{2}{3}b\right)^2\)
c,\(a^2+10ab+25b^2\)=\(\left(a+5b\right)^2\)
d,\(x^2-6xy^2+9y^4\) =(\(x-3y^2\))\(^2\)
e,\(a^2+a+\frac{1}{4}\)=\(\left(a+\frac{1}{2}\right)^2\)
Cho mik điểm SP đi ak <3
\(a,\left(x+2y\right)^2=x^2+4xy+4y^2\)
\(b,\left(3x-2y\right)^2=9x^2-12xy+4y^2\)
\(c,\left(2x-\dfrac{1}{2}\right)^3=8x^3-3.4x^2.\dfrac{1}{2}+3.2x.\dfrac{1}{4}-\dfrac{1}{8}=8x^3-6x^2+\dfrac{3}{2}x-\dfrac{1}{8}\)
\(d,\left(\dfrac{x}{2}-y\right)\left(\dfrac{x}{2}+y\right)=\dfrac{x^2}{4}-y^2\)
\(2;a,x^4+4x^2+4\)
\(=\left(x^2+2\right)^2\)
\(b,4a^2b^2-c^2d^2\)
\(=\left(2ab\right)^2-\left(cd\right)^2\)
\(=\left(2ab-cd\right)\left(2ab+cd\right)\)
Bài1:
\(\left(3+xy^2\right)^2=81+6xy^2+x^2y^4\)
Các câu sau tương tự
Bài2:
\(a,\left(4x^2+4xy+y^2\right)\)
=\(\left(2x+y\right)^2\)
b)\(9m^2+n^2-6mn=\left(3m-n\right)^2\)
c)\(16a^2+25b^2+40ab=\left(4a+5b\right)^2\)
d)\(x^2-x+\dfrac{1}{4}=\left(x-\dfrac{1}{2}\right)^2\)
Bài3:
\(a,301^2=\left(300+1\right)^2=900+600+1=1501\)
b/\(499^2=\left(500-1\right)^2=2500-1000+1=1501\)
c/\(68.72=\left(70-2\right)\left(70+2\right)=70^2-2^2=4900-4=4896\)
Bài 1:
a, \(\left(3+xy^2\right)^2=9+6xy^2+x^2y^4\)
b, \(\left(10-2m^2n\right)^2=100-40m^2n+4m^4n^2\)
c, \(\left(a-b\right)^2.\left(a+b\right)^2=\left[\left(a-b\right)\left(a+b\right)\right]^2\)
\(=\left(a^2-b^2\right)^2=a^4-2a^2b^2+b^4\)
Chúc bạn học tôt!!!